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Ta có:
\(0,4\left(3\right)=\frac{43-4}{90}=\frac{39}{90}=\frac{13}{30}.\)
\(0,6\left(2\right)=\frac{62-6}{90}=\frac{56}{90}=\frac{28}{45}.\)
\(0,6\left(8\right)=\frac{68-6}{90}=\frac{62}{90}=\frac{31}{45}.\)
Vậy:
\(\frac{13}{30}+\frac{28}{45}.\frac{5}{2}-\frac{\frac{5}{6}}{\frac{31}{45}}.\frac{53}{50}\)
\(=\frac{13}{30}+\frac{14}{9}-\frac{75}{62}.\frac{53}{50}\)
\(=\frac{13}{30}+\frac{14}{9}-\frac{159}{124}\)
\(=\frac{179}{90}-\frac{159}{124}\)
\(=\frac{3943}{5580}.\)
Chúc bạn học tốt!
\(0,4\left(3\right)=\frac{43-4}{90}=\frac{39}{90}=\frac{13}{30}\)đó Michiel Girl Mít ướt
Đó là công thức đưa 1 số thập phân vô hạn tuần hoàn sang phân số đó Michiel Girl Mít ướt
\(=\dfrac{13}{30}+\dfrac{28}{45}\cdot\dfrac{5}{2}\cdot\left[\dfrac{5}{6}:\dfrac{53}{90}\right]\cdot\dfrac{53}{50}\)
\(=\dfrac{13}{30}+\dfrac{14}{9}\cdot\dfrac{5}{6}\cdot\dfrac{90}{53}\cdot\dfrac{53}{50}\)
\(=\dfrac{13}{30}+\dfrac{7}{3}=\dfrac{83}{30}\)
Ta có: \(0,4\left(3\right)=\dfrac{43-4}{90}=\dfrac{39}{90};\) \(0,6\left(2\right)=\dfrac{62-6}{90}=\dfrac{56}{90}\)
\(0,5\left(8\right)=\dfrac{58-5}{90}=\dfrac{53}{90}\)
Vậy biểu thức M có giá trị:
\(\dfrac{39}{90}+\dfrac{56}{90}.\dfrac{5}{2}-\dfrac{\dfrac{5}{6}}{\dfrac{53}{90}}.\dfrac{53}{30}=\dfrac{13}{30}+\dfrac{14}{9}-\dfrac{5}{6}.\dfrac{90}{53}.\dfrac{53}{50}=\dfrac{13}{30}+\dfrac{14}{9}-\dfrac{3}{2}\)
\(=\dfrac{13.9+14.30-3.135}{270}=\dfrac{402}{270}=\dfrac{67}{45}\)
Tính:
0,4(3) + 0,6(2) . \(2\frac{1}{2}\).[(\(\frac{1}{2}+\frac{1}{3}\)) : 0,5(8)] : \(\frac{50}{53}\)
0,4(3) + 0,6(2). \(2\frac{1}{2}\).\(\left[\left(\frac{1}{2}+\frac{1}{3}\right):0,5\left(8\right)\right]:\frac{50}{53}\)
\(=\frac{13}{30}+\frac{28}{45}.\frac{5}{2}.\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)
\(=\frac{13}{30}+\frac{28}{45}.\frac{5}{2}.\frac{75}{53}:\frac{50}{53}\)
\(=\frac{13}{30}+\frac{7}{3}\)
\(=\frac{83}{30}\)
\(A=0,4\left(3\right)+0,6\left(2\right)\cdot2\frac{1}{2}-\frac{\frac{1}{2}+\frac{1}{3}}{0,5\left(8\right)}:\frac{50}{53}\)
\(A=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{3+2}{6}:\frac{53}{90}\cdot\frac{53}{50}\)
\(A=\frac{13}{30}+\frac{14}{9}-\frac{5}{6}\cdot\frac{90}{53}\cdot\frac{53}{50}\)
\(A=\frac{39}{90}+\frac{140}{90}-\frac{2}{3}\)
\(A=\frac{179}{90}-\frac{60}{90}=\frac{119}{90}\)
\(A=1,3\left(2\right)\)